What is (x+1)^2+(y+2)^2=8 ?

The solution to (x+1)^2+(y+2)^2=8 is Circle with (a,b)=(-1,-2),r=sqrt(8)

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(x+1)^2+(y+2)^2=8

(x + 1)^{2} + (y + 2)^{2} = 8

(a, b) = (- 1, - 2), r = 8

Solution steps

(x + 1)^{2} + (y + 2)^{2} = 8

Rewrite

(x + 1)^{2} + (y + 2)^{2} = 8

in the form of the standard circle equation

(x - (- 1))^{2} + (y - (- 2))^{2} = (8)^{2}

Therefore the circle properties are:

(a, b) = (- 1, - 2), r = 8

x^{2} + y^{2} + 12 x + 18 y + 92 = 0

x^{2} + y^{2} - 2 y = 3

x^{2} + y^{2} - 10 x - 10 y + 46 = 0

x^{2} + y^{2} - 16 y - 36 = 0

3 x^{2} + 3 y^{2} - 24 x + 6 y - 24 = 0

What is (x+1)^2+(y+2)^2=8 ?
The solution to (x+1)^2+(y+2)^2=8 is Circle with (a,b)=(-1,-2),r=sqrt(8)